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Idealist
Group Title
If we intercept an electron having total energy 1533 MeV that came from Vega, which is 26 ly from us, how far in lightyears was the trip in the rest frame of the electron?
 2 years ago
 2 years ago
Idealist Group Title
If we intercept an electron having total energy 1533 MeV that came from Vega, which is 26 ly from us, how far in lightyears was the trip in the rest frame of the electron?
 2 years ago
 2 years ago

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Carl_Pham Group TitleBest ResponseYou've already chosen the best response.0
You need to work backward from the relativistic momentumenergy formula to find the velocity of the electron in our reference frame, then use that with the LorentzFitzgerald contraction to find the ratio between the distance in the electron's frame and in ours.
 2 years ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.0
Here's the first part: http://hyperphysics.phyastr.gsu.edu/hbase/relativ/releng.html
 2 years ago

gleem Group TitleBest ResponseYou've already chosen the best response.1
From the special theory of relativity \[t' =t/\sqrt{1v ^{2}/c ^{2}}\]where t is the time in the electron's reference frame and t' is the time in our reference frame and v is the velocity we observe . Also from the special theory we have\[E=mc ^{2}=m _{0}c ^{2}/\sqrt{1v ^{2}/c ^{2}}\]where \[m _{0}\]is the rest mass of the object . The rest mass of an electron is .511MeV Solve for t in terms of v,E and mo.
 2 years ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
mc^2=8.187*10^14 J (8.187*10^14)/1533=5.34*10^17 What do I do after that?
 2 years ago

gleem Group TitleBest ResponseYou've already chosen the best response.1
Because energy can be expressed in electron volts and mass can be expressed as energy (E=m0*c^2) you do not have to change to joules and kgms in this problem. You can if you want but you must work with the rest mass in kgm the velocity of light in m/sec and energy in joules. what do we need to do first. We want the time in the moving reference frame of the electron,t. We have the equation that relates that time to the observer time on earth, t'. we know c. So we must find the velocity of the electron with an total energy of 1533 MeV. Actually we only need to find (v/c)^2. since that quantity appears in the two equations we are using. Since we have the total energy of the electron (i.e. rest mass energy +KE) E=1533MeV we can find its velocity. with the equation\[E=mc ^{2}=m _{0}c ^{2}/\sqrt{1(v/c)^{2}}\] E=mc^2 =.1533MeV solve for v/c and substitute that into the time equation to find t. remember m0c^2=0.511Mev and t'=27 years.
 2 years ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
I get it now. Thanks.
 2 years ago
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